Partial differential equation help I have a linear pde $ 3U_{xx} -2U_{xt} -U_{tt} = 0   ,           -\infty < x > \infty$ 
I have solved this and found the general solution which is 
$ U(x,t) = F(x-3t) + G(x+t) $ 
G ,F  are arbitrary twice differentiable functions of their argument.
I am stuck on the second part of this question which askes how to solve 
Show that the solution satisfying
$ U(x,0) = f(x) , U_{t}(x,0) = g(x) $
is  $  u(x,t) =  \frac{1}{4} [ f(x-3t) + 3f(x+t) + \int_{x-3t}^{x+t} g(\xi) d\xi]$
I know I have to make use of this equation 
$ u(x,t) = \frac{1}{2} [\varphi(x-ct) + \varphi(x+ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} \si)(\xi)d\xi $ 
But I am really stuck on proving this, any help appreciated. 
thank you. 
 A: Your formula with $x\pm ct$ seems more adapted to solve an equation like $U_{xx}-U_{tt}=0$.
Here, you are almost over. Start with the general form of a solution :
$$U(x,t)=F(x-3t)+G(x+t)$$
Now, to use the first boundary condition, just set $t=0$ in the above formula. You get :
$$F(x)+G(x)=f(x) \; \; \; (1)$$
Then, derive the expression of $U$ with respect to $t$ : $U_t(x,t)=G'(x+t)-3F'(x-3t)$. Then, set $t=0$ and use the second boundary equation to get :
$$G'(x)-3F'(x)=g(x) \; \; \; (2)$$
Deriving equation (1), you have :
$$F'(x)+G'(x)=f'(x) \; \; \; (3)$$
Combining equations (2) and (3) (just a little linear algebra), you find $F'$ and $G'$ :
$$\left\lbrace \begin{array}{l}
F'(x) = \frac{1}{4}(f'(x)-g(x)) \\
G'(x) = \frac{1}{4}(g(x)+3f'(x))
\end{array}\right. \; \; \; \; \; \; (4)$$
We define h as : $h(x)=\int_0^x g(\xi) \text{d}\xi$. Integrating (4), one get :
$$\left\lbrace \begin{array}{l}
F(x) = \frac{1}{4}(f(x)-h(x))+A \\
G(x) = \frac{1}{4}(h(x)+3f(x))+B
\end{array}\right. \; \; \; \; \; \; (5)$$
with $A,B$ integration constants
Injecting thes expression in the formula for $U$ (with $C=A+B$) :
$$U(x,t)=\frac{1}{4} \left(f(x-3t)-h(x-3t)+h(x+t)+3f(x+t)\right) + C$$
We have $h(x+t)-h(x-3t)=\int_0^{x+t} g(\xi)\text{d}\xi-\int_0^{x-3t} g(\xi) \text{d}\xi=\int_{x-3t}^{x+t} g(\xi) \text{d}\xi$, so :
$$U(x,t)=\frac{1}{4} \left(f(x-3t)+3f(x+t)\right) + \frac{1}{4} \int_{x-3t}^{x+t} g(\xi) \text{d}\xi + C$$
Finally, take $t=0$, you find $C=0$ with the first boundary condition, so :
$$\boxed{U(x,t)=\frac{1}{4} \left[ f(x-3t)+3f(x+t) + \int_{x-3t}^{x+t} g(\xi) \text{d}\xi \right]}$$
